Find integral $\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$ I have to find definite integration $$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$$ 
I tried to subtitute $x = \sin u$  and $x = e^u$ 
but got no result . Please help me in proceeding.
 A: The substitution 
$$x = e^u$$
is perfect, if you know how to proceed.
$$x = e^u ~~~~~ \text{d}x = e^{u}\ \text{d}u$$
Hence you get
$$\int_{-\infty}^0 \frac{u e^{u}}{1 - e^{2u}}\ \text{d}u$$
We recognize the possibility to use the Geometric Series with the fraction
$$\frac{1}{1 - e^{2u}} = \sum_{k = 0}^{+\infty} e^{2uk}$$
So
$$\int_{-\infty}^0\ ue^{u}\ \sum_{k = 0}^{+\infty} e^{2uk}\ \text{d}u = \sum_{k = 0}^{+\infty}\int_{-\infty}^0\ u\ e^{u(1 + 2k)}\ \text{d}u$$
Now the integration is trivial if you compute it by parts with
$$f = u ~~~~~ g' = e^{u(1 + 2k)}$$
Integrating by parts one and you get finally
$$\sum_{k = 0}^{+\infty} \frac{-1}{(1 + 2k)^2} = -\frac{\pi^2}{8}$$
By parts calculation
$$\int_{-\infty}^0\ u\ e^{u(1 + 2k)}\ \text{d}u = \frac{u}{1+2k}e^{u(1+2k)}\bigg|_{-\infty}^0 - \int_{-\infty}^0\ \frac{e^{u(1 + 2k)}}{1+2k}\ \text{d}u$$
The first term is zero, hence you need to evaluate the trivial integral
$$ - \int_{-\infty}^0\ \frac{e^{u(1 + 2k)}}{1+2k}\ \text{d}u = -\frac{1}{(1+2k)^2}e^{u(1+2k)}\bigg|_{-\infty}^0 = \frac{-1}{(1+2k)^2}$$
A: $$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$$
Note that for $|x|<1$:
$$\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}$$
Hence for $|x^2|<1$:
$$\sum_{n=0}^{\infty} x^{2n}=\frac{1}{1-(x^2)}$$
So we get:
$$=\int_{0}^{1} \sum_{n=0}^{\infty} x^{2n}\log x dx$$
$$=\sum_{n=0}^{\infty} \int_{0}^{1} x^{2n} \log x  dx$$
Integration by parts.
$$=\sum_{n=0}^{\infty} -\frac{1}{(2n+1)^2}$$
$$=-\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$$
Now think about evens and odds to see that this is equivalently:
$$=-\left( \sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2n)^2}\right)$$
$$=-\frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
A: The first step is to turn the one integral into two as follows:
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = \frac{1}{2} \, \left[ \int_{0}^{1} \frac{\ln(x)}{1-x} \, dx + \int_{0}^{1} \frac{\ln(x)}{1+x} \, dx \right]
\end{align}
then using the integrals
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x} \, dx &= \left[Li_{2}(1-x) \right]_{0}^{1} = - \zeta(2) = - \frac{\pi^2}{6} \\
\int_{0}^{1} \frac{\ln(x)}{1+x} \, dx &= - \frac{\pi^2}{12}
\end{align}
the desired integral becomes
$$\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = - \frac{\pi^2}{8}.$$
A: Per Feynman’s trick
$$I(a)
=\frac12 \int_0^1 \frac{\ln (1+a^2x^2)-\ln(1+a^2) }{1 - x^2} dx
$$
$$I’(a)= -\frac {1}{1+a^2}\int_0^1\frac {a dx}{1+a^2x^2}=-\frac{\tan^{-1}a}{1+a^2}
$$
Then
\begin{align}
\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x
& = I(\infty) = \int_0^\infty I’(a)da =
-\int_0^\infty \frac{\tan^{-1}a}{1+a^2}da
=-\frac{\pi^2}8
\end{align}
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\Psi}$ is the Digamma Function:
\begin{align}
\color{#f00}{\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x} &=
\left.\partiald{}{\mu}\int_{0}^{1}{x^{\mu} - 1 \over 1 - x^{2}}\,\dd x
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] & \stackrel{x\ \to\ x^{1/2}}{=}\,\,\,\,\,
\left.\half\,\partiald{}{\mu}
\int_{0}^{1}{x^{\mu/2 - 1/2}\,\,\,\, -\,\, x^{-1/2} \over 1 - x}\,\dd x
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] & =
\half\,\partiald{}{\mu}
\bracks{\Psi\pars{\half} - \Psi\pars{{\mu \over 2} + \half}}_{\ \mu\ =\ 0}
\\[5mm] & =
-\,{1 \over 4}\,\Psi\,'\pars{\half} =
-\,{1 \over 4}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1/2}^{2}} =
-\sum_{k = 0}^{\infty}{1 \over \pars{2k + 1}^{2}}
\\[5mm] & =
-\sum_{k = 0}^{\infty}{1 \over k^{2}} +
\sum_{k = 0}^{\infty}{1 \over \pars{2k}^{2}} =
-\,{3 \over 4}\
\underbrace{\sum_{k = 0}^{\infty}{1 \over k^{2}}}_{\ds{\pi^{2} \over 6}}\ =\
\color{#f00}{-\,{\pi^{2} \over 8}} 
\end{align}
